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CHEMISTRY- boiling point

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0.1m aqueous solution of HCl and 0.1m aqueous solution of C6H12O6.
calculate the boiling point of each

i know the eq. dTb=Kbm
but i have no temperatures. this is the question

  • CHEMISTRY- boiling point - ,

    It is an aqueous solution. You know that the boiling point with no HCl is 100 C. Use Kb and the molarity to compute the change in B.P.

  • CHEMISTRY- boiling point - ,

    can you show me. i'm confused.

  • CHEMISTRY- boiling point - ,

    delta T = i*Kb*molality
    i is the van't Hoff factor, which for HCl is 2 (it dissolves into two particles).
    delta T = 2*0.512*0.1 = 0.102
    Therefore, the new boiling point for water must be 100 C + 0.102 = ??

    For the sugar solution,
    i = 1 since it doesn't ionize.
    delta T = i*Kb*m
    delta T = 1*0.512*0.1 = 0.0512
    New boiling point will be
    100 C + 0.0512.
    The problem is hoping you will see the difference between the two; i.e., the HCl ionizes to give two particles and has just twice the effect as the other one although the molality is the same.

  • CHEMISTRY- boiling point - ,

    is the Kb 0.51 just standard BP of water? i think that's where i'm getting lost.

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